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Mathematics LibreTexts

4.S: Mathematical Induction (Summary)

  • Page ID
    7058
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    Important Definitions

    • Inductive set, page 171

    • Factorial, page 201

    • Recursive definition, page 200

    • Fibonacci numbers, page 202

    • Geometric sequence, page 206

    • Geometric series, page 206

    The Various Forms of Mathematical Induction

    1. The Principle of Mathematical Induction
      If \(T\) is a subset of \(\mathbb{N}\) such that 

      (a) \(1 \in T\), and 
      (b) For every \(k \in \mathbb{N}\), if \(k \in T\), then \((k + 1) \in T\).

      then \(T = \mathbb{N}\)
      Procedure for a Proof by Mathematical Induction
      To prove \((\forall n \in \mathbb{N}\)(P(n))\)
      \[\begin{array} {rcl} {\text{Basis step}} &: & {\text{Prove} P(1).} \\ {\text{Inductive step}} &: & {\text{Prove that for each} k \in \mathbb{N}, \text{if} P(k) \text{is true, then} P(k + 1) \text{is true.}} \end{array}\]

    2. The Extended Principle of Mathematical Induction
      Let \(M\) be an integer. If \(T\) is a subset of \(\mathbb{Z}\) such that

      (a) \(M \in T\), and
      (b) For every \(k \in \mathbb{Z}\) with \(k \ge M\), if \(k \in T\), then \((k + 1) \in T\).

      then \(T\) contains all integers greater than or equal to \(M\).
      Using the Extended Principle of Mathematical Induction
      Let \(M\) be an intteger. To prove \((\forall n \in \mathbb{Z} \text{with} n \ge M) (P(n))\)
      \[\begin{array} {rcl} {\text{Basis step}} &: & {\text{Prove} P(M).} \\ {\text{Inductive step}} &: & {\text{Prove that for each} k \in \mathbb{Z} \text{with} k \ge M, \text{if} P(k) \text{is true, then} P(k + 1) \text{is true.}} \end{array}\]
      We can then conclude that \(P(n)\) is true for all \(n \in \mathbb{Z}\) with \(n \ge M\).

    3. The Second Principle of Mathematical Induction
      Let \(M\) be an integer. If \(T\) is a subset of \(\mathbb{Z}\) such that

      (a) \(M \in T\), and
      (b) For every \(k \in \mathbb{Z}\) with \(k \ge M\), if \(\{M, M + 1, ..., k\} \subseteq T\), then \((k + 1) \in T\).

      then \(T\) contains all integers greater than or equal to \(M\).
      Using the Second Principle of Mathematical Induction
      Let \(M\) be an intteger. To prove \((\forall n \in \mathbb{Z} \text{with} n \ge M) (P(n))\)
      \[\begin{array} {rcl} {\text{Basis step}} &: & {\text{Prove} P(M).} \\ {\text{Inductive step}} &: & {\text{Let} k \in \mathbb{Z} \text{with} k \ge M. \text{Prove that if} P(M), P(M + 1), ..., P(k) \text{are true, then} P(k + 1) \text{is true.}} \end{array}\]
      We can then conclude that \(P(n)\) is true for all \(n \in \mathbb{Z}\) with \(n \ge M\).

    Important Results

    • Theorem 4.9. Each natural number greater than 1 is either a prime number or is a product of prime numbers.

    • Theorem 4.14. Let \(a, r \in \mathbb{R}\). If a geometric sequence is defined by \(a_1 = a\) and for each \(n \in \mathbb{N}\), \(a_{n + 1} = r \cdot a_n\), then for each \(n \in \mathbb{N}\), \(a_n = a \cdot r^{n - 1}\).

    • Theorem 4.15. Let \(a, r \in \mathbb{R}\). If the sequence \(S_1, S_2, ..., S_n, ...\) is defined by \(S_1 = a\) and for each \(n \in \mathbb{N}\), \(S_{n + 1} = a + r \cdot S_n\), then for each \(n \in \mathbb{N}\), \(S_n = a + a \cdot r + a \cdot r^2 + \cdot\cdot\cdot + a \cdot r^{n - 1}\). That is, the geometric series \(S_n\) is the sum of the first n terms of the corresponding geometric sequence.

    • Theorem 4.16. Let \(a, r \in \mathbb{R}\) and \(r \ne 1\). If the sequence \(S_1, S_2, ..., S_n, ...\) is defined by \(S_1 = a\) and for each \(n \in \mathbb{N}\), \(S_{n + 1} = a + r \cdot S_n\), then for each \(n \in \mathbb{N}\), \(S_n = a (\dfrac{1 - r^n}{1 - r})\).